Why Frey's curve cannot be constructed? My question is related to the fact that Frey's curve:
$$y^{2} = x(x-a^{p})(x+b^{p})$$
Could not be constructed if Fermat's last theorem holds true. I mean because Fermat's last theorem implies that there is no solution for $a$, $b$, $c$, and prime number $p > 2$ such that:
$$a^{p}+b^{p}=c^{p}$$ 
As a result the Frey's curve which is semi-stable elliptic curve could be not constructed because according to Ribet's theorem this curve is not modular but Taniyama-Shimura-Weil conjecture implies that each elliptic curve should be modular. The thing I can't understand that is when we say Frey's curve could not be constructed in what sense it could not be constructed? I mean the equation $y^{2} = x(x-a^{p})(x+b^{p})$ contains only $a^{p}$ and $b^{p}$ but it's not completely symmetric to contain all $a$, $b$, $c$ in a symmetric way. I appreciate if someone could explain this to me.
 A: In 1969 Hellegouarch performed the elliptic curves $F (a, b)$, which were later named after Gerhard Frey and were constructed from the solutions of the Fermat equation, he reported on the curves in an article in Acta Arithmetica. I have found the article but it is not in English.
Let me give you a over-simplified version taken form the lecture-series "Riemann Hypothesis and its Applications" (Lecture-29) by Prof. Manindra Agrawal, Department of Computer Science and Engineering, IIT Kanpur (won the 2006 Fulkerson Prize, and the 2006 Gödel Prize for the AKS primality test).
First Step
Let $F:= y^2=x^3+ax^2+bx+c$, then discriminant $\Delta_F=(ab)^2+4a^3c-4b^3+18abc-27c^3.$
If $c=0$, then $y^2=x^3+ax^2+bx$, $\Delta_F=(ab)^2-4b^3.$
Now the Frey's curve is , $y^2=x(x-A)(x+B)=(x^2-Ax)(x+B)=x^3-Ax^2+Bx^2-ABx.$
Let, $a=(B-A), b=-AB,$ then $\Delta= \{(B-A)(-AB)\}^2-4(-AB)^3= \{(B-A)^2-4(-AB)\}(-AB)^2=(A^2+B^2-2AB+4Ab)(AB)^2=(A+B)^2(AB)^2.$
Now, let, $A=a^q, B=b^q,$ then, $\Delta_F=(a^q+b^q)^2(ab)^{2q},$
if $a^q+b^q=c^q$ is true for $q>2$ (this is Fermat's last theorem),
then the discriminant $\Delta_F=(c^q)^2(ab)^{2q}=(abc)^{2q}.$
Second Step
If $\Delta_F$ is $l^{th}$ power of an integer, then it has a point of order $l$. If $a^q+b^q=c^q$ is true for $q>2$, then the Frey curve $F$ has a point a point of order $2q$, since it's discriminant $\Delta_F$ is $2q^{th}$ power of an integer $(abc)$.
Third Step
But if $F$ is modular, then, it does not have a point of order $ > C$ where $C$ is a constant.
Fourth Step
It has been proved that every elliptic curve is modular (Taniyama–Shimura conjecture, Andrew Wiles proved the Taniyama–Shimura conjecture for semistable elliptic curves, which was enough to imply Frey curve is modular).
Final Step
Since Frey curve $F$ is modular, then it can not have a point of order $ \geq C$, but the Frey curve $F$'s discriminant $\Delta_F$ is $2q^{th}$ power of an integer $(abc)$, so $q$ can not be $> \frac{C}{2}$, does there could not be solution $a^q+b^q=c^q$ for $q> \frac{C}{2}$, i.e. there are finite $q$ for which Fermat's Last Theorem is true.
Frey's curve can not be constructed since by the definition of the Frey's Curve, it must have a point of order $2q$ but at the same time it is an elliptic curve, thus modular, so, the Frey's Curve can not have a point of order $ \geq C$, so after certain value of $q$, Frey's curve can not be constructed.
Of-course I have missed many intricate details for the sake of accessibility.
In a bit more technical terms: The group of $p$-torsion points $F[p]$ on $F$
(which is a two-dimensional vector space over the field $\mathbb F_p$ of $p$-elements),
equipped with an action of the Galois group of $\overline{\mathbb Q}$ over $\mathbb Q$)  has very special properties,
in algebraic number theory terms, it is very close to being unramified, specifically, but more technically, it is unramified except possibly at $2$ and $p$, and at $p$ the ramification is very mild, it is finite flat.
Now the Shimura-Taniyama conjecture, which is what Wiles (together with Taylor)
proved, shows that $F$, and so $F[p]$, arises from a weight two modular form.  Ribet's earlier results on Serre's epsilon conjecture imply that this modular
form must actually be of level $2$. (This is where we use the above information about the ramification.)   But there are no non-zero cuspforms of weight $2$ and level $2$, and we get a contradiction.
Although it is much harder (in that the only way we know to rule out the existence of $F[p]$ is by the - quite difficult- Shimura-Taniyama
conjecture, or else by related more recent results such as Khare and Wintenberger's work on Serre's conjecture), one can think of the non-existence
of $F[p]$ as being analogous to Minkowski's theorem in algebraic number theory, which says that an everywhere-unramified extension of $\mathbb Q$ cannot exist (See this answer).
For detail proof, read following books:
1. Fermat's Last Theorem: Basic Tools by Takeshi Saito
(Publication Year: 2013),
2. Fermat's Last Theorem: The Proof by Takeshi Saito (Publication Year: 2014).
